操作系统 课后答案(7)

v(wmutex);

p(rwmutex); //其他写进程在rwmutex上排队

write data; //临界区

v(rwmutex); p(wmutex); --writecount;

if(writecount==0) v(x); //写进程都写完时，通过v(x)允许读进程读 v(wmutex); }

附加题：

读者优先，规定仅允许5个进程同时读，怎样修改程序？

解：增加一个资源信号量s，初值为5。 int s=5;

Reader：

begin P(rmutex); readcount=readcount+1; if(readcount==1)then P(rwmutex); V(rmutex); P(s); read_file(); V(s); P(rmutex); readcount=readcount-1; if(readcount==0)then V(rwmutex); V(rmutex); end

writer: